thôi bỏ hình đê
\(\Delta MBO\sim\Delta M'B'O\)\(\Leftrightarrow\dfrac{MB}{M'B'}=\dfrac{BO}{B'O}\Leftrightarrow\dfrac{h}{h1}=\dfrac{d_1}{d_1'}\Rightarrow d_1=\dfrac{d_1'h}{h_1}\left(1\right)\)
\(\Delta OIF\sim\Delta B'M'F'\Leftrightarrow\dfrac{OI}{M'B'}=\dfrac{OF'}{B'F'}\Leftrightarrow\dfrac{h}{h_1}=\dfrac{f}{d_1'+f}\)
\(\Rightarrow d_1'=\dfrac{f\left(h_1-h\right)}{h}\left(2\right)\)
Tương tự:\(d_2=\dfrac{d_2'h}{h_2}\left(3\right)\) \(\Rightarrow d_2'=\dfrac{f\left(h_2-h\right)}{h}\left(4\right)\)
\(d_3=\dfrac{d_3'h}{h_3}\left(5\right)\)\(\Rightarrow d_3'=\dfrac{f\left(h_3-h\right)}{h}\left(6\right)\)
Theo bài ra ta có:
(1)(2)(3) \(d_1+d_2=2d_3\)
\(\Leftrightarrow\dfrac{d_1'h}{h_1}+\dfrac{d_2'h}{h_2}=\dfrac{2d_3'h}{h_3}\)
\(\Leftrightarrow\dfrac{d_1'}{h_1}+\dfrac{d_2'}{h_2}=\dfrac{2d_3'}{h_3}\)
\(\Leftrightarrow\dfrac{f\left(h_1-h\right)}{h_1h}+\dfrac{f\left(h_2-h\right)}{h_2h}=\dfrac{2f\left(h_3-h\right)}{h_3h}\)
\(\Leftrightarrow\dfrac{h_1-h}{h_1}+\dfrac{h_2-h}{h_2}=\dfrac{2h_3-2h}{h_3}\)
\(\Leftrightarrow1-\dfrac{h}{h_1}+1-\dfrac{h}{h_2}=2-\dfrac{2h}{h_3}\)
\(\Leftrightarrow\dfrac{h}{h_1}+\dfrac{h}{h_2}+\dfrac{2h}{h_3}\)
\(\Leftrightarrow\dfrac{2}{h_3}=\dfrac{1}{h_1}+\dfrac{1}{h_2}\)
=>\(h_3=2\left(cm\right)\)
:)) maybe ko chắc
